Examples

Chapter 6 Class 12 Application of Derivatives
Serial order wise

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### Transcript

Example 33 Find intervals in which the function given by f(π₯) =3/10 π₯4 β 4/5 π₯^3β 3π₯2 + 36/5 π₯ + 11 is (a) strictly increasing (b) strictly decreasingf(π₯) = 3/10 π₯4 β 4/5 π₯^3β 3π₯2 + 36/5 π₯ + 11 Finding fβ(π) fβ(π₯) = 3/10 Γ 4π₯^3 β 4/5 Γ 3π₯^2 β 3 Γ 2x + 36/5 + 0 fβ(π₯) = 12/10 π₯^3β 12/5 π₯^2β 6x + 36/5 fβ(π₯) = 6/5 π₯^3β 12/5 π₯^2β 6x + 36/5 fβ(π₯) = 6(π₯^3/5β(2π₯^2)/5βπ₯+6/5) fβ(π₯) = 6((π₯^3 β 2π₯^2β 5π₯ + 6)/5) = 6/5 (π₯^3β2π₯^2β5π₯+6) = 6/5 (π₯β1)(π₯2βπ₯β6) = 6/5 (π₯β1)(π₯2β3π₯+2π₯β6) = 6/5 (π₯β1)[π₯(π₯β3)+2(π₯β3)] = 6/5 (π₯β1)(π₯+2)(π₯β3) Hence, fβ(π) = π/π (πβπ)(π+π)(πβπ) Putting fβ(π) = 0 π/π (πβπ)(π+π)(πβπ) = 0 (π₯β1)(π₯+2)(π₯β3) = 0 Hence, x = β2 , 1 & 3 Plotting points on number line Hence, f(π₯) is strictly decreasing on the interval π₯ β (ββ,βπ)& (π , π) f(π₯) is strictly increasing on the interval π₯ β (βπ,π) & (π , β)